Pair Of Straight Lines Question 84
Question: The angle between the lines represented by the equation $ ax^{2}+2hxy+by^{2}=0 $ is given by
[RPET 1995]
Options:
A) $ \tan \theta =\frac{2(h^{2}-ab)}{(a+b)} $
B) $ \tan \theta =\frac{2\sqrt{h^{2}-ab}}{(a+b)} $
C) $ \tan \theta =\frac{2(h^{2}-ab)}{\sqrt{a+b}} $
D) $ \tan \theta =\frac{2\sqrt{h^{2}+ab}}{(a+b)} $
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Answer:
Correct Answer: B
Solution:
Let m_1 and m_2 be the slopes of the lines by the equation
Given, $ax^2+2hxy+by^2=0$ ……. (i)
Let $y=m_1x$ and $y=m_2x$
$\therefore(y−m_1x)(y−m_2x)=0$
and $m_1m_2x^2−(m_1+m_2)xy+y^2=0$ ……… (ii)
Comparing (i) and (ii), we get
$\frac{m_1m_2}{a}=\frac{1}{b}=\frac{m_1+m_2}{2h}$
$∴m_1m_2=\frac{a}{b} and m_1+m_2=\frac{−2h}{b}$
Thus, $(m_1−m_2)^2=(m_1+m_2)^2−4m_1m_2$
$(m_1−m_2)^2=(\frac{−2h}{b})^2−4(\frac{a}{b})$
$(m_1−m_2)^2=4(h^2−ab)b^2$
Let the angle between $ y=m_1x$ and$ y=m_2x$ be θ.
$ \tan \theta =\frac{2\sqrt{h^{2}-ab}}{(a+b)} $ if a+b≠0
BETA
